[Math] Binomial Calculating Probability of Airline Tickets

binomial distributionstatistics

Because not all airline passengers show up for their reserved seat, an airline sells 125 tickets for a flight that holds only 115 passengers. The probability that a passenger does not show up is 0.05, and the passengers behave independently. Round your answers to four decimal places (e.g. 98.7654).

a) What is the probability that every passenger who shows up can take the flight?

b) What is the probability that the flight departs with at least one empty seat?

I am not using a statistics program to calculate my answers like I have seen many answers on here use, but by using a formula:
For example: P(115) =
125 C 115 * (.95)^115 * (.05)^10 = .0475

Similarly I have:
P(116) = .0778
P(117) = .1137
P(118) = .1465
P(119) = .1637
P(120) = .1556
P(121) = .1221
P(122) = .0761
P(123) = .0353
P(124) = .0108
P(125) = .0016

So for part a) I did:
1 – P(X > 115) = 1 – .9032 = .0968

and for part b) I did:
1 – P(x >= 115) = 1 – .9507 = .0493

These numbers just do not seem correct to me. And I am confused as to why the probabilites are increasing from P(115) to P(119), (I would expect them to decrease, however I guess if they are on the rising part of the binomial distribution and then go to the falling part of the distribution at P(120)

Edit: I know understand these values are correct and fall around the most probable value P(119) which is the mean. Thank you for help in clarifying.

Best Answer

The number of people who show up for the flight in the circumstances you describe is $X \sim \mathsf{Binom}(n=125, p=.95).$ The probability everyone who shows has a seat is $P(X \le 115) \approx 0.0967.$ This can be computed in R statistical software, in which pbinom is a binomial CDF.

pbinom(115, 125, .95)
[1] 0.09672946

This may seem absurdly small, but the average number of people showing is $\mu = E(X)= np = 128.25 > 125.$ So on the average flight the airline should expect to leave some people behind.

The probability there will be at least one empty seat is $P(X \le 114) = 0.0492.$ So that will seldom happen.

pbinom(114, 125, .95)
[1] 0.04921917

Here is a plot of the relevant part of the distribution of $X.$

enter image description here

Here are exact values from R, matching your computations, which seem quite accurate:

x = 115:125;  pdf = dbinom(x, 125, .95)
cbind(x, pdf)
        x         pdf
 [1,] 115 0.047510291
 [2,] 116 0.077818581
 [3,] 117 0.113734849
 [4,] 118 0.146505907
 [5,] 119 0.163741896
 [6,] 120 0.155554801
 [7,] 121 0.122129803
 [8,] 122 0.076080861
 [9,] 123 0.035256984
[10,] 124 0.010804560
[11,] 125 0.001642293

Note: Code for figure:

x = 110:125;  pdf = dbinom(x, 125, .95)
plot(x, pdf, type="h", ylim=c(0,max(pdf)), lwd=2)
abline(h=0, col="green3")
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